42v^2+96v+24=0

Solution for 42v^2+96v+24=0 equation:

42v^2+96v+24=0

a = 42; b = 96; c = +24;
Δ = b2-4ac
Δ = 962-4·42·24
Δ = 5184
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5184}=72$
$v_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-72}{2*42}=\frac{-168}{84}
=-2
$
$v_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+72}{2*42}=\frac{-24}{84}
=-2/7
$